Answer to Question #194255 in Calculus for Maureen

Question #194255

Find the slope of the tangent line at the incident point

3e to the power x = xy+xsquared+ysquared at (0,1)

Write an equation of the tangent line


1
Expert's answer
2021-05-19T08:30:50-0400
(3e)x=xy+x2+y2(3e)^x=xy+x^2+y^2

Differentiate both parts with respect to xx . Use the Chain Rule


ddx((3e)x)=ddx(xy+x2+y2)\dfrac{d}{dx}((3e)^x )=\dfrac{d}{dx}(xy+x^2 +y^2)

(ln3+1)(3e)x=y+xdydx+2x+2ydydx(\ln3+1)(3e)^x=y+x\dfrac{dy}{dx}+2x+2y\dfrac{dy}{dx}

Solve for dydx\dfrac{dy}{dx}


dydx=(ln3+1)(3e)x2xyx+2y\dfrac{dy}{dx}=\dfrac{(\ln3+1)(3e)^x-2x-y}{x+2y}



Find the slope of the tangent line at point (0,1)(0, 1)


slope=m=dydx(0,1)slope=m=\dfrac{dy}{dx}|_{(0,1)}

=(ln3+1)(3e)02(0)10+2(1)=ln32=\dfrac{(\ln3+1)(3e)^0-2(0)-1}{0+2(1)}=\dfrac{\ln3}{2}

The equation of the tangent line in point-slope form


y1=ln32(x0)y-1=\dfrac{\ln3}{2}(x-0)

The equation of the tangent line in slope-intercept form


y=ln32x+1y=\dfrac{\ln3}{2}\cdot x+1


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment